2.8 is 4% of what number

Accounting Question

Romashka [77]

B inventory
Add purchase
Add fright inward

Total amount to be account for
Less E inventory

7 0

3 years ago

Sqrt(32x^3) - sqrt(16x^3) + root(4,x^3) -sqrt(2x^3)

almond37 [142]

An expression is defined as a set of numbers, variables, and mathematical operations. The simplified version of the given expression is (3√2-2)√x³.

<h3>What is an Expression?</h3>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

The given expression $\sqrt{32x^3} - \sqrt{16x^3} + \sqrt{4x^3} -\sqrt{2x^3}$ can be simplified as shown below.

$\sqrt{32x^3} - \sqrt{16x^3} + \sqrt{4x^3} -\sqrt{2x^3}\\\\=\sqrt{x^3}(\sqrt{32}-\sqrt{16}+\sqrt{4}-\sqrt{2}})\\\\= \sqrt{x^3} (4\sqrt{2}-4+2-\sqrt2)\\\\= \sqrt{x^3} (3\sqrt{2}-2)$

Hence, The simplified version of the given expression is (3√2-2)√x³.

Learn more about Expression:

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7 0

2 years ago

The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit

Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b) We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean $\mu$ , standard deviation $\sigma$

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

8 0

3 years ago

I just need help solving this problem. Also please don't put the answer into a file because I can't open them. It's also really

Pavlova-9 [17]

Answer:

The value of x is 27

because 2 times 27 equals 54 minus 7 equals 47

5 0

3 years ago

Find thetwo consecutive integers whose product is 24

malfutka [58]

Answer: